So the chem II....

<p>anybody wanna quiz me, i haven't studied yet and i'm thinkin about starting maybe. (i'm a soph so it doesn't really matter, i'll just cancel if i suck)</p>

<p>do you have a prep book?</p>

<p>Alright, I'll start: </p>

<p>A weak acid does NOT have which of the following characteristics:</p>

<p>I. does not dissociate at all in aqueous solution
II. Ka<0
III. accepts an electron pair</p>

<p>A. I only
B. III only
C. II and III only
D. I and II only
E. I, II, and III</p>

<p>E? 10 char</p>

<p>that NOT really confused me with I. haha
Is the answer D? if not can you explain the whole Ka and Kb thing to me</p>

<p>the answer is indeed D. </p>

<p>Does a weak acid not dissociate at all? Of course it dissociates, even if it's only a limited degree.</p>

<p>Does a weak acid have a Ka<0? Nope... Ka is the concentration of ions (H+ and A-) over the concentration of molecules of acid. Concentration is never negative, Ka is never negative.
[When K is less than 1, there are more reactants, and when K is greater 1, there are more products. But for strong acids, Ka is infinity, because there are virtually no molecules of acid not dissociated. Kb is just the same expression, but for a base.] </p>

<p>Does a weak acid accept an electron pair? Yes... all acids do.</p>

<p>Therefore D.</p>

<p>Msu, do you need more explanation on Ka/Kb?</p>

<p>well, I just don't really understand what its a measurement of.
and isn't pH= pKa +log [conjugate acid or A-]/[conjugate base or HA] for buffers? And... Kesp=Ka?? and what is pKa (the p part of it)</p>

<p>pKa is the -log Ka.</p>

<p>The equation you posted, the Henderson Hasselbalch (sp?) equation, is indeed for buffers. </p>

<p>Ksp is solubility product. [precipitate] <--> [cation][anion]</p>

<p>Ka and Kb are for acids and bases only.</p>

<p>EDIT: oops didn't see first sentence. Ka, like any other equilibrium constant, is the ratio of products to reactants. In acids, the ratio of H+/A- ions to HA molecules.</p>

<p>So the conjugate acid and A- are the same as well as the conjugate base and HA? and can the concentration of HA be replaced with the concentration of the salt if the base isn't given?</p>

<p>Thank you so much. You've cleared up so much</p>

<p>The conjugate base, of the acid HA, is A-. I don't quite understand what you're asking.</p>

<p>Eq. expression is </p>

<p>



     [H+][A-]
Ka = --------
       [HA]


</p>

<p>And you're welcome. Glad to hear that some things are cleared up.</p>

<p>so aren't they 2 different equations? one with conjgate acid/conjugate base and the other is the equation you wrote above?</p>

<p>oh yeah, the handerson- "hasselhuff" equation. yeah there are two different one. one for pH, one for pOH.</p>

<p>I'll try to add a question.</p>

<p>which of the following shows the wavelength of electromagnetic waves in INCREASING order?</p>

<p>a. x-ray, gamma ray, infrared, microwave
b. visible, infrared, microwave, radio wave
c, ultraviolet, visible, x-ray, gamma
d, radiowave, visible, ultraviolet, x-ray.</p>

<p>Is it B? :/</p>

<p>B</p>

<p>For some reason, chemistry counts microwaves separately from radio waves. Oh well.</p>

<p>Anybody got a really challenging one??</p>

<p>wait WHAT WAS THAT?! i have never seen that before and I read 2 of the books all the way through</p>

<p>I'm pretty sure Barron's had a diagram in it with the wavelengths of electromagnetic radiations?</p>

<p>oh. thats the one i didnt read. well is it complicated or important for the test because barrons tends to go into a lot of detail</p>

<p>I'll try to add a question.</p>

<p>which of the following shows the wavelength of electromagnetic waves in INCREASING order?</p>

<p>a. x-ray, gamma ray, infrared, microwave
b. visible, infrared, microwave, radio wave
c, ultraviolet, visible, x-ray, gamma
d, radiowave, visible, ultraviolet, x-ray.</p>

<p>Don't think that'll be tested, but it's B. (oh wait, that was already answered)</p>

<p>I'll add a question</p>

<p>How much HCl must be added to neutralize 0.01 L of 1M Ca(OH)2?</p>

<p>A. 0.001 mol
B. 0.01 mol
C. 0.02 mol
D. 0.03 mol
E. 1.00 mol</p>

<p>Use stoichiometry (hard to write on this but let me try)</p>

<p>0.01L x (1 mol Ca(OH)2 / 1000 mL ) x (2 mol OH- / 1 mol Ca(OH)2) x (1 mol H+ / 1 mol OH-) x (1 mol HCl / 1 mol H+ ) </p>

<p>All this should = 0.02</p>

<p>yeah since theres an 2OH then just multiply its M by 2</p>

<p>wow that's easier lol</p>